Respuesta :
Answer:
a) tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex] \bar X= \frac{3.2+2.0+2.5+5.0}{4}=3.175[/tex]
b) [tex] \bar X = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}[/tex]
And replacing we got:
[tex] \bar X= \frac{3.2*6+2.0*3+2.5*2+5.0*8}{6+3+2+8}=\frac{70.2}{19}=3.695[/tex]
Step-by-step explanation:
Part a
For this case we can calculate the the unweigthed with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex] \bar X= \frac{3.2+2.0+2.5+5.0}{4}=3.175[/tex]
Part b
For this case the weigthed mena is given :
[tex] \bar X = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}[/tex]
And replacing we got:
[tex] \bar X= \frac{3.2*6+2.0*3+2.5*2+5.0*8}{6+3+2+8}=\frac{70.2}{19}=3.695[/tex]
